Question

Consider the following plots of log of rate constant $ k (log k)$ vs $ \frac{1}{T} $ for three different reactions. The correct order of activation energies of these reactions is:

Choose the correct answer from the options given below:

• A. \( Ea_2>Ea_1>Ea_3 \)
• B. \( Ea_1>Ea_3>Ea_2 \)
• C. \( Ea_1>Ea_2>Ea_3 \)
• D. \( Ea_3>Ea_2>Ea_1 \)

Hint:

The slope of the Arrhenius plot is directly related to the activation energy. A steeper slope corresponds to a higher activation energy.

Answer 1

The Correct Option is A

The activation energy \( E_a \) of a reaction is related to the slope of the plot of \( \log k \) vs \( \frac{1}{T} \) by the Arrhenius equation: \[ \log k = -\frac{E_a}{2.303R} \times \frac{1}{T} + \text{constant} \] where: - \( \log k \) is the log of the rate constant, - \( T \) is the temperature, - \( R \) is the gas constant. The steeper the slope, the higher the activation energy. In this plot: - Reaction 1 (the line with the steepest slope) corresponds to the highest activation energy, i.e., \( E_{a1} \). - Reaction 2 (the line with a less steep slope) corresponds to the middle activation energy, i.e., \( E_{a2} \). - Reaction 3 (the line with the least steep slope) corresponds to the lowest activation energy, i.e., \( E_{a3} \).
Thus, the correct order of activation energies is \( E_{a2}>E_{a1}>E_{a3} \), which corresponds to option
(1)
.